Question

The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth house, then

• A. 2 $\le$ a $\le$ 6
• B. 8 $\le$ a $\le$ 12
• C. 14 $\le$ a $\le$ 20
• D. 22 $\le$ a $\le$ 30

Answer 1

The correct option is C

14 $\le$ a $\le$ 20

Let the house numbers be $\alpha ,\alpha +2,\alpha +4,\alpha +6,\alpha +8,\alpha +10\dots ..$

$\alpha +10=a\Rightarrow \alpha =a–10$ …….(1)

House no. will be (+)

$\Rightarrow \alpha =a–10>0$

$\Rightarrow \alpha =a–10>0$

$\Rightarrow \alpha >10$

$\Rightarrow \alpha \ge$ 12 as a is each too........... (2)

$Now,S_n=\frac{n}{2}[2\alpha +(n+1\left)d\right]$

$170=\frac{n}{2}[2\alpha +(n+1\left)2\right]$

$=n\alpha +(n+1))$

$=n(a-10+n-1)$

$=n(a-11+n)$

$\Rightarrow n^2+n(a-11)-170=0$

$\Rightarrow n=\frac{(11-a)\pm \sqrt{(a-11{)}^2+680}}{2}.....\left(3\right)$

$\because n\ge 6$

$\Rightarrow \frac{(11-a)\pm \sqrt{(a-11{)}^2+680}}{2}\ge 6$

$\Rightarrow a\le \frac{800}{24}.......\left(4\right)$

From (2) and (4) $\Rightarrow 12\le a\le 32$

Now checking through (3) for a = 12, 14...

we have a = 18, n = 10 and $S_n$ = 170